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3 years ago

2) If the current in any circuit reach to infinity then its resistance becomes

Physics

1 answer:

In-s [12.5K]3 years ago

4 0

Answer:

Zero according to ohm's law R= V / i

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According to the kinetic theory, collisions between molecules in a gas _____.a.are inelasticb.never occurc. cause a loss of tota

KonstantinChe [14]

Answer:

d. perfectly elastic

Explanation:

According to the kinetic theory for collisions of gas molecules:

1.The loss of energy is negligible or we can say that it is zero.

2.Molecules of the gas move in a random manner.

3.The collision between molecules and with the wall of the container is perfectly elastic.That is why loss in the energy is zero.

Therefore the correct answer will be d.

d. perfectly elastic

3 0

3 years ago

I need helpppppp asap

asambeis [7]

It’s c because it’s not Control so that means that it would be broken and non fix able

5 0

3 years ago

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

3 years ago

Please help me answer this

ivann1987 [24]

Answer:

A is the correct answer

Hope it helps

4 0

2 years ago

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con

ira [324]

The Period of the resulting shm will be T=39.7

Explanation:

Given data

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

By newton's second law.

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

using the k=mω value

k=mω

ω=k/m

ω=1200/3

ω=400

Using F = -kx value

x = F/-k

x=(-10.16)/1200

x=0.00847m

Restoring the torque value

τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

Rotational to linear form

a = αr

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

Time Period

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0

3 years ago